YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 35 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 9 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QDP
        (10) UsableRulesProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (18) YES
    (19) QDP
        (20) UsableRulesProof [EQUIVALENT, 0 ms]
        (21) QDP
        (22) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (23) YES
    (24) QDP
        (25) UsableRulesProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (28) YES
    (29) QDP
        (30) UsableRulesProof [EQUIVALENT, 0 ms]
        (31) QDP
        (32) QDPOrderProof [EQUIVALENT, 33 ms]
        (33) QDP
        (34) PisEmptyProof [EQUIVALENT, 0 ms]
        (35) YES


----------------------------------------

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   r(r(x1)) -> s(r(x1))
   r(s(x1)) -> s(r(x1))
   r(n(x1)) -> s(r(x1))
   r(b(x1)) -> u(s(b(x1)))
   r(u(x1)) -> u(r(x1))
   s(u(x1)) -> u(s(x1))
   n(u(x1)) -> u(n(x1))
   t(r(u(x1))) -> t(c(r(x1)))
   t(s(u(x1))) -> t(c(r(x1)))
   t(n(u(x1))) -> t(c(r(x1)))
   c(u(x1)) -> u(c(x1))
   c(s(x1)) -> s(c(x1))
   c(r(x1)) -> r(c(x1))
   c(n(x1)) -> n(c(x1))
   c(n(x1)) -> n(x1)

Q is empty.

----------------------------------------

(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   r(r(x1)) -> r(s(x1))
   s(r(x1)) -> r(s(x1))
   n(r(x1)) -> r(s(x1))
   b(r(x1)) -> b(s(u(x1)))
   u(r(x1)) -> r(u(x1))
   u(s(x1)) -> s(u(x1))
   u(n(x1)) -> n(u(x1))
   u(r(t(x1))) -> r(c(t(x1)))
   u(s(t(x1))) -> r(c(t(x1)))
   u(n(t(x1))) -> r(c(t(x1)))
   u(c(x1)) -> c(u(x1))
   s(c(x1)) -> c(s(x1))
   r(c(x1)) -> c(r(x1))
   n(c(x1)) -> c(n(x1))
   n(c(x1)) -> n(x1)

Q is empty.

----------------------------------------

(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(b(x_1)) = x_1
   POL(c(x_1)) = x_1
   POL(n(x_1)) = 1 + x_1
   POL(r(x_1)) = 1 + x_1
   POL(s(x_1)) = x_1
   POL(t(x_1)) = x_1
   POL(u(x_1)) = 1 + x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   r(r(x1)) -> r(s(x1))
   n(r(x1)) -> r(s(x1))
   u(r(t(x1))) -> r(c(t(x1)))
   u(n(t(x1))) -> r(c(t(x1)))




----------------------------------------

(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   s(r(x1)) -> r(s(x1))
   b(r(x1)) -> b(s(u(x1)))
   u(r(x1)) -> r(u(x1))
   u(s(x1)) -> s(u(x1))
   u(n(x1)) -> n(u(x1))
   u(s(t(x1))) -> r(c(t(x1)))
   u(c(x1)) -> c(u(x1))
   s(c(x1)) -> c(s(x1))
   r(c(x1)) -> c(r(x1))
   n(c(x1)) -> c(n(x1))
   n(c(x1)) -> n(x1)

Q is empty.

----------------------------------------

(5) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(r(x1)) -> R(s(x1))
   S(r(x1)) -> S(x1)
   B(r(x1)) -> B(s(u(x1)))
   B(r(x1)) -> S(u(x1))
   B(r(x1)) -> U(x1)
   U(r(x1)) -> R(u(x1))
   U(r(x1)) -> U(x1)
   U(s(x1)) -> S(u(x1))
   U(s(x1)) -> U(x1)
   U(n(x1)) -> N(u(x1))
   U(n(x1)) -> U(x1)
   U(s(t(x1))) -> R(c(t(x1)))
   U(c(x1)) -> U(x1)
   S(c(x1)) -> S(x1)
   R(c(x1)) -> R(x1)
   N(c(x1)) -> N(x1)

The TRS R consists of the following rules:

   s(r(x1)) -> r(s(x1))
   b(r(x1)) -> b(s(u(x1)))
   u(r(x1)) -> r(u(x1))
   u(s(x1)) -> s(u(x1))
   u(n(x1)) -> n(u(x1))
   u(s(t(x1))) -> r(c(t(x1)))
   u(c(x1)) -> c(u(x1))
   s(c(x1)) -> c(s(x1))
   r(c(x1)) -> c(r(x1))
   n(c(x1)) -> c(n(x1))
   n(c(x1)) -> n(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 7 less nodes.
----------------------------------------

(8)
Complex Obligation (AND)

----------------------------------------

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   N(c(x1)) -> N(x1)

The TRS R consists of the following rules:

   s(r(x1)) -> r(s(x1))
   b(r(x1)) -> b(s(u(x1)))
   u(r(x1)) -> r(u(x1))
   u(s(x1)) -> s(u(x1))
   u(n(x1)) -> n(u(x1))
   u(s(t(x1))) -> r(c(t(x1)))
   u(c(x1)) -> c(u(x1))
   s(c(x1)) -> c(s(x1))
   r(c(x1)) -> c(r(x1))
   n(c(x1)) -> c(n(x1))
   n(c(x1)) -> n(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(10) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   N(c(x1)) -> N(x1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(12) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*N(c(x1)) -> N(x1)
The graph contains the following edges 1 > 1


----------------------------------------

(13)
YES

----------------------------------------

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   R(c(x1)) -> R(x1)

The TRS R consists of the following rules:

   s(r(x1)) -> r(s(x1))
   b(r(x1)) -> b(s(u(x1)))
   u(r(x1)) -> r(u(x1))
   u(s(x1)) -> s(u(x1))
   u(n(x1)) -> n(u(x1))
   u(s(t(x1))) -> r(c(t(x1)))
   u(c(x1)) -> c(u(x1))
   s(c(x1)) -> c(s(x1))
   r(c(x1)) -> c(r(x1))
   n(c(x1)) -> c(n(x1))
   n(c(x1)) -> n(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(15) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   R(c(x1)) -> R(x1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(17) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*R(c(x1)) -> R(x1)
The graph contains the following edges 1 > 1


----------------------------------------

(18)
YES

----------------------------------------

(19)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(c(x1)) -> S(x1)
   S(r(x1)) -> S(x1)

The TRS R consists of the following rules:

   s(r(x1)) -> r(s(x1))
   b(r(x1)) -> b(s(u(x1)))
   u(r(x1)) -> r(u(x1))
   u(s(x1)) -> s(u(x1))
   u(n(x1)) -> n(u(x1))
   u(s(t(x1))) -> r(c(t(x1)))
   u(c(x1)) -> c(u(x1))
   s(c(x1)) -> c(s(x1))
   r(c(x1)) -> c(r(x1))
   n(c(x1)) -> c(n(x1))
   n(c(x1)) -> n(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(20) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(21)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(c(x1)) -> S(x1)
   S(r(x1)) -> S(x1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(22) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*S(c(x1)) -> S(x1)
The graph contains the following edges 1 > 1


*S(r(x1)) -> S(x1)
The graph contains the following edges 1 > 1


----------------------------------------

(23)
YES

----------------------------------------

(24)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U(s(x1)) -> U(x1)
   U(r(x1)) -> U(x1)
   U(n(x1)) -> U(x1)
   U(c(x1)) -> U(x1)

The TRS R consists of the following rules:

   s(r(x1)) -> r(s(x1))
   b(r(x1)) -> b(s(u(x1)))
   u(r(x1)) -> r(u(x1))
   u(s(x1)) -> s(u(x1))
   u(n(x1)) -> n(u(x1))
   u(s(t(x1))) -> r(c(t(x1)))
   u(c(x1)) -> c(u(x1))
   s(c(x1)) -> c(s(x1))
   r(c(x1)) -> c(r(x1))
   n(c(x1)) -> c(n(x1))
   n(c(x1)) -> n(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(25) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(26)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U(s(x1)) -> U(x1)
   U(r(x1)) -> U(x1)
   U(n(x1)) -> U(x1)
   U(c(x1)) -> U(x1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(27) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*U(s(x1)) -> U(x1)
The graph contains the following edges 1 > 1


*U(r(x1)) -> U(x1)
The graph contains the following edges 1 > 1


*U(n(x1)) -> U(x1)
The graph contains the following edges 1 > 1


*U(c(x1)) -> U(x1)
The graph contains the following edges 1 > 1


----------------------------------------

(28)
YES

----------------------------------------

(29)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   B(r(x1)) -> B(s(u(x1)))

The TRS R consists of the following rules:

   s(r(x1)) -> r(s(x1))
   b(r(x1)) -> b(s(u(x1)))
   u(r(x1)) -> r(u(x1))
   u(s(x1)) -> s(u(x1))
   u(n(x1)) -> n(u(x1))
   u(s(t(x1))) -> r(c(t(x1)))
   u(c(x1)) -> c(u(x1))
   s(c(x1)) -> c(s(x1))
   r(c(x1)) -> c(r(x1))
   n(c(x1)) -> c(n(x1))
   n(c(x1)) -> n(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(30) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(31)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   B(r(x1)) -> B(s(u(x1)))

The TRS R consists of the following rules:

   u(r(x1)) -> r(u(x1))
   u(s(x1)) -> s(u(x1))
   u(n(x1)) -> n(u(x1))
   u(s(t(x1))) -> r(c(t(x1)))
   u(c(x1)) -> c(u(x1))
   s(r(x1)) -> r(s(x1))
   s(c(x1)) -> c(s(x1))
   r(c(x1)) -> c(r(x1))
   n(c(x1)) -> c(n(x1))
   n(c(x1)) -> n(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(32) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   B(r(x1)) -> B(s(u(x1)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(B(x_1)) = 2*x_1
   POL(c(x_1)) = 0
   POL(n(x_1)) = 0
   POL(r(x_1)) = 1 + 4*x_1
   POL(s(x_1)) = x_1
   POL(t(x_1)) = 1 + 4*x_1
   POL(u(x_1)) = x_1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   u(r(x1)) -> r(u(x1))
   u(s(x1)) -> s(u(x1))
   u(n(x1)) -> n(u(x1))
   u(s(t(x1))) -> r(c(t(x1)))
   u(c(x1)) -> c(u(x1))
   s(r(x1)) -> r(s(x1))
   s(c(x1)) -> c(s(x1))
   r(c(x1)) -> c(r(x1))
   n(c(x1)) -> c(n(x1))
   n(c(x1)) -> n(x1)


----------------------------------------

(33)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   u(r(x1)) -> r(u(x1))
   u(s(x1)) -> s(u(x1))
   u(n(x1)) -> n(u(x1))
   u(s(t(x1))) -> r(c(t(x1)))
   u(c(x1)) -> c(u(x1))
   s(r(x1)) -> r(s(x1))
   s(c(x1)) -> c(s(x1))
   r(c(x1)) -> c(r(x1))
   n(c(x1)) -> c(n(x1))
   n(c(x1)) -> n(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(34) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(35)
YES